I have a question regarding the definition of stopping times in continuous and discrete time.
For continuous time we have the definition that a random variable on a filtered probability space $(\Omega , \mathcal{F}, (\mathcal{F}_{n})_{n \in I}, \mathbb{P})$ is called stopping time if
$\lbrace \tau \leq n \rbrace \in \mathcal{F}_{n} \:\: \forall n \in I$.
Now if time $I$ is dicrete you can show that the definition above is equivalent to
$\lbrace \tau = n \rbrace \in \mathcal{F}_{n} \:\: \forall n \in I$
My question now is: Why does the definition that works in dicrete time doesn't work in continuous time?
In continuous time, you can have a non-trivial random variables $\tau$ with $\mathbb{P}(\tau = t) = 0$ for all $t$. In addition, in continuous time, it is common to augment $\mathcal F_0$ by all of the $\mathbb{P}$-null sets of $\mathcal F_\infty$. This would create problems with allowing things that really shouldn't be stopping times to be stopping times if the definition we took was "$\{\tau = t\} \in \mathcal F_t$ for all $t$." For example, if $B_t$ is a Brownian motion, the random variable $\tau := |B_1|$ would satisfy $\mathbb{P}(\tau = t) = 0$ for all $t$, and hence $\{\tau = t\} \in \mathcal F_0$ (assuming we are taking the augmented filtration). In particular, this implies $\{\tau = t\} \in \mathcal F_t \supset \mathcal F_0$ for all $t$, so $|B_1|$ would be a stopping time under this definition. However, we clearly don't want $|B_1|$ to be a stopping time, so we can't take this as the definition of a stopping time.